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Ex 3.6, 1 (v) and (vi) - 7x - 2y / xy = 5, 8x + 7y / xy = 15 - Ex 3.6

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise
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Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v) (7๐‘ฅ โˆ’ 2๐‘ฆ)/๐‘ฅ๐‘ฆ = 5 (8๐‘ฅ +7๐‘ฆ)/๐‘ฅ๐‘ฆ = 15 Given (7๐‘ฅ โˆ’ 2๐‘ฆ)/๐‘ฅ๐‘ฆ = 5 (7๐‘ฅ )/๐‘ฅ๐‘ฆ - (2๐‘ฆ )/๐‘ฅ๐‘ฆ = 5 (7 )/๐‘ฆ โˆ’(2 )/๐‘ฅ = 5 โˆ’(2 )/๐‘ฅ +(7 )/๐‘ฆ = 5 Our equations are โˆ’(2 )/๐‘ฅ +(7 )/๐‘ฆ = 5 โ€ฆ(1) (7 )/๐‘ฅ +(8 )/๐‘ฆ = 15 ...(2) Hence, we solve โ€“2u + 7v = 5 โ€ฆ(3) 7u + 8v = 15 โ€ฆ(4) From (3) โ€“2u + 7v = 5 7v = 5 + 2u v = (5 + 2๐‘ข)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2๐‘ข)/7) = 15 Multiplying 7 both sides 7 ร— 7u + 7 ร— 8((5+2๐‘ข)/7) = 7 ร— 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 โ€“ 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) โ€“2u + 7v = 5 โ€“2(1) + 7v = 5 โ€“2 + 7v = 5 7v = 5 + 2 7v = 7 v = 7/7 v = 1 Hence u = 1 , v = 1 But we have to find x & y We know that u = 1/๐‘ฅ 1 = 1/๐‘ฅ x = 1 Hence, x = 1 , y = 1 is the solution of the given equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (vi) 6x + 3y = 6xy 2x + 4y = 5xy Given 6x + 3y = 6xy Diving whole equation by xy (6๐‘ฅ+3๐‘ฆ)/๐‘ฅ๐‘ฆ = 6๐‘ฅ๐‘ฆ/๐‘ฅ๐‘ฆ 6๐‘ฅ/๐‘ฅ๐‘ฆ +3๐‘ฆ/๐‘ฅ๐‘ฆ = 6 6/๐‘ฆ +3/๐‘ฅ = 6 Hence, our equations are 6/๐‘ฆ +3/๐‘ฅ = 6 โ€ฆ(1) 2/๐‘ฆ +4/๐‘ฅ = 5 โ€ฆ(2) Now, we solve 6v + 3u = 6 โ€ฆ(3) 2v + 4u = 5 โ€ฆ(4) From (3) 6v + 3u = 6 6v = 6 โ€“ 3u v = (6 โˆ’ 3๐‘ข)/6 Putting value of v in (4) 2v + 4u = 5 2((6 โˆ’3๐‘ข)/6) + 4u = 5 ((6 โˆ’3๐‘ข)/3) + 4u = 5 Multiplying both sides by 3 3 ร— ((6 โˆ’4๐‘ข)/3) + 3 ร— 4u = 3 ร— 5 (6 โ€“ 3u) + 12u = 15 โ€“3u + 12u = 15 โ€“ 6 9u = 9 u = 9/9 u = 1 Putting u = 1 in (3) 6v + 3u = 6 6v + 3(1) = 6 6v + 3 = 6 6v = 6 โ€“ 3 6v = 3 v = 3/6 v = 1/2 Hence u = 1 , v = 1/2 But we have to find x & y Now, u = 1/๐‘ฅ 1 = 1/๐‘ฅ x = 1 Hence, x = 1 , y = 2 is the solution of the given equation

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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